# If #2^x xx4^(x+1)=8# what is the value of #x#?

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High detail using first principles. Plus an alternative approach for the end.

Split each side in half.

You can solve the above problem directly, but I'd like to show you another way that might work in a different situation:

divide each side by three.

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To find the value of ( x ), we can start by simplifying the equation ( 2^x \times 4^{x+1} = 8 ). Since ( 4 ) can be expressed as ( 2^2 ), we can rewrite ( 4^{x+1} ) as ( (2^2)^{x+1} ). Using the property of exponents, we get ( 4^{x+1} = 2^{2(x+1)} ). Substituting this back into the original equation, we have ( 2^x \times 2^{2(x+1)} = 8 ). Now, using the property of exponents for multiplication, we add the exponents when the bases are the same, resulting in ( 2^{x+2x+2} = 8 ). Simplifying the exponent, we have ( 2^{3x+2} = 8 ). Since ( 8 = 2^3 ), we can rewrite the equation as ( 2^{3x+2} = 2^3 ). Setting the exponents equal to each other, we get ( 3x+2 = 3 ). Solving for ( x ), we subtract ( 2 ) from both sides to get ( 3x = 1 ), then divide both sides by ( 3 ) to find ( x = \frac{1}{3} ). So, the value of ( x ) is ( \frac{1}{3} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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